$2\cosec A$
The task is to simplify the following trigonometric expression:
\( \frac{\sin A}{1-\cos A} + \frac{1-\cos A}{\sin A} \)
Identify the common denominator for the two fractions, which is \((1-cos A)sin A\).
Rewrite the expression with the common denominator:
\( \frac{\sin^2 A}{(1-\cos A)\sin A} + \frac{(1-\cos A)^2}{\sin A (1-\cos A)} \)Combine the numerators over the common denominator:
\( \frac{\sin^2 A + (1-\cos A)^2}{(1-\cos A)\sin A} \)Expand the squared term in the numerator: $(1-cos A)2 = 1 - 2cos A + cos2 A$.
\( \frac{\sin^2 A + 1 - 2\cos A + \cos^2 A}{(1-\cos A)\sin A} \)Use the Pythagorean identity $sin2 A + cos2 A = 1$.
Substitute \(1\) for $sin2 A + cos2 A$:
\( \frac{1 + (1 - 2\cos A)}{(1-\cos A)\sin A} \)Simplify the numerator:
\( \frac{2 - 2\cos A}{(1-\cos A)\sin A} \)Factor out \(2\) from the numerator:
\( \frac{2(1 - \cos A)}{(1-\cos A)\sin A} \)Cancel the common factor \((1 - cos A)\), assuming \( A \neq 2n\pi \):
\( \frac{2}{\sin A} \)Apply the reciprocal identity \(cosec A = 1/sin A\):
\( 2\cosec A \)The simplified expression is \(2\cosec A\).
What is the simplified value of the expression \(\frac{\cos A}{1-\tan A}+\frac{\sin A}{1-\cot A}\) ?
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